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The 2010, Honorary Member Welcoming, Undergraduate Degree Results Chat Thread Mk. IV watch

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    Can the ones who know maths well check my working on this question and see if I've done it right, and what I can do next please?
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    (Original post by Loz17)
    Can the ones who know maths well check my working on this question and see if I've done it right, and what I can do next please?
    Which question?
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    (Original post by maxfire)
    Oooh. Take it. Then you can dive me places.

    Ermm..not sure exactly. ask Chis.


    YO. UPDATES ON THE MEET?:love:
    6 people responded to the form this time.
    A_master, flowermaster91, maxfire, CrazyChris, Smeh, Loz17

    From these, there were no dates with 0 'no' responses, but five with 1 'no'
    In best order, these are
    19th June - 3 yes, 2 unsure, 1 no.
    26th June - 3 yes, 2 unsure, 1 no.
    Then:
    20th March - 2 yes, 3 unsure, 1 no.
    10th April - 2 yes, 3 unsure, 1 no.
    And
    17th April - 1 yes, 4 unsure, 1 no.

    So, take from that what you will.
    More people should definitely fill the thread meet availability form out. (hint Jenni, +/-, silent and others!)
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    (Original post by maxfire)
    Which question?
    \frac{3x^2 + x + 2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)}

    3x^2 + x + 2 = A(x+1)(x) + B(x+1) + C(x^2)

    If thats correct, what could I sub in for x to eliminate 2 variables? Was thinking x = -1 to find C and then try and equate coefficients to find A and B (but I'm not so good at that)
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    The top bit. Why does 2 become 1?

    And you would use 0 and -1, I think.
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    (Original post by Loz17)




    If thats correct, what could I sub in for x to eliminate 2 variables? Was thinking x = -1 to find C and then try and equate coefficients to find A and B (but I'm not so good at that)
    -1 :yes:
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    (Original post by CrazyChris)
    6 people responded to the form this time.
    A_master, flowermaster91, maxfire, CrazyChris, Smeh, Loz17

    From these, there were no dates with 0 'no' responses, but five with 1 'no'
    In best order, these are
    19th June - 3 yes, 2 unsure, 1 no.
    26th June - 3 yes, 2 unsure, 1 no.
    Then:
    20th March - 2 yes, 3 unsure, 1 no.
    10th April - 2 yes, 3 unsure, 1 no.
    And
    17th April - 1 yes, 4 unsure, 1 no.

    So, take from that what you will.
    More people should definitely fill the thread meet availability form out. (hint Jenni, +/-, silent and others!)
    I'm sure I filled that form in. :dunce:
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    (Original post by + polarity -)
    The top bit. Why does 2 become 1?

    And you would use 0 and -1, I think.
    Just a typo. Thanks.
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    Thanks all
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    (Original post by sil3nt_cha0s)
    oh wait, do I have to do it in a more general sense?

    this is where I'm going wrong methinks :facepalm2: you can't just do it for something like u(t) = Cx^2 + Dx + E, can you? cos that's too specific, amirite?
    Yes, because you have to show that if u'' is not zero then you have no solution.

    What you are doing are just testing specific cases (when there are infact an infinite number of functions u(t))
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    (Original post by CrazyChris)
    6 people responded to the form this time.
    A_master, flowermaster91, maxfire, CrazyChris, Smeh, Loz17

    From these, there were no dates with 0 'no' responses, but five with 1 'no'
    In best order, these are
    19th June - 3 yes, 2 unsure, 1 no.
    26th June - 3 yes, 2 unsure, 1 no.
    Then:
    20th March - 2 yes, 3 unsure, 1 no.
    10th April - 2 yes, 3 unsure, 1 no.
    And
    17th April - 1 yes, 4 unsure, 1 no.

    So, take from that what you will.
    More people should definitely fill the thread meet availability form out. (hint Jenni, +/-, silent and others!)
    I filled it in ages ago :ta:
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    \frac{3x^2 + x + 2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)}

    3x^2 + x + 2 = A(x+1)(x) + B(x+1) + C(x^2)

    If thats correct, what could I sub in for x to eliminate 2 variables? Was thinking x = -1 to find C and then try and equate coefficients to find A and B (but I'm not so good at that)
    One final problem. I got C=4 and A=2. Now how can I find B?
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    :hmmm:
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    (Original post by Loz17)
    \frac{3x^2 + x + 2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)}

    3x^2 + x + 2 = A(x+1)(x) + B(x+1) + C(x^2)

    If thats correct, what could I sub in for x to eliminate 2 variables? Was thinking x = -1 to find C and then try and equate coefficients to find A and B (but I'm not so good at that)
    To equate the coeffecients, use what you have... expand the brackets.

    3x^2 +x +2 = Ax^2 + Ax + Bx + B + Cx^2
    3x^2 +x +2 = (A+C)x^2 + (A + B)x + B

    Take each coefficient in turn:
    x^2: 3 = A + C
    x: 1 = A + B
    0: 2 = B

    Then from there, just solve the simultaneous equations. (substitue B = 2 into the middle. then subsitute your value for A into the first)
    Hope this helps. [Probably late..!]
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    (Original post by CrazyChris)
    To equate the coeffecients, use what you have... expand the brackets.

    3x^2 +x +2 = Ax^2 + Ax + Bx + B + Cx^2
    3x^2 +x +2 = (A+C)x^2 + (A + B)x + B

    Take each coefficient in turn:
    x^2: 3 = A + C
    x: 1 = A + B
    0: 2 = B

    Then from there, just solve the simultaneous equations. (substitue B = 2 into the middle. then subsitute your value for A into the first)
    Hope this helps. [Probably late..!]
    Actually thats solved my final problem. I needed to find B as I've got A and C by substitution. Thanks
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    Looks good :yep:

    Well, put the values of A and C into the equation, and sub x as 0
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    (Original post by Loz17)
    One final problem. I got C=4 and A=2. Now how can I find B?
    I got A = -1
    hmm.
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    (Original post by maxfire)
    :hmmm:
    :hugs:
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    Dammit too late :hmmm:

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    (Original post by CrazyChris)
    I got A = -1
    hmm.
    Yea I see where I went wrong. B=2 not A
 
 
 
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